MATH 332 - Exponential Functions Question
Bank
Question 1
Problem Statement:
The population of a small island was 1,500 in the year 2010 and has been
growing exponentially since then. By the year 2015, the population had grown
to 2,450. Assume the growth can be modeled by an exponential function of the
form P(t) = P0·ekt, where P(t) represents the population at year t,P0is the
initial population at the starting time, and kis the rate of growth.
1. Find the exponential growth function that models the population of the
island. 2. Predict the population of the island in the year 2023.
Step-by-Step Solution:
Step 1: Use given data to find the growth rate k.
Given: - P(0) = 1500 (Population at the year 2010) - P(5) = 2450 (Popula-
tion at the year 2015, where t= 5 since 2015 - 2010 = 5)
The equation is:
2450 = 1500 ·e5k
Firstly solve for e5k:
e5k=2450
1500 ≈1.6333
Now, solve for k:
5k= ln(1.6333)
k=ln(1.6333)
5≈0.4908
5= 0.09816
Step 2: Write the growth function.
Using the initial population and the growth rate, the function becomes:
P(t) = 1500 ·e0.09816t
Step 3: Predict the population for the year 2023.
To find the population in 2023: - t= 2023 −2010 = 13
Substitute t= 13 into the population function:
P(13) = 1500 ·e0.09816×13
1
P(13) = 1500 ·e1.27608
P(13) = 1500 ·3.582
P(13) ≈5373
Conclusion:
The population model for the island is P(t) = 1500 ·e0.09816t, and the pre-
dicted population in the year 2023 is approximately 5,373.
Note: This question provides a comprehensive start to understanding expo-
nential functions and their applications in real-world scenarios. It encourages
students to manipulate exponential equations and use logarithms, which are
part of the curriculum at Liberty University. Question 1: Understanding
Growth through Exponential Functions
Problem Statement:
The population of a small island was 1,500 in the year 2010 and
has been growing exponentially since then. By the year 2015, the
population had grown to 2,450. Assume the growth can be modeled
by an exponential function of the form P(t) = P0·ekt, where P(t)
represents the population at year t,P0is the initial population at the
starting time, and kis the rate of growth.
1. Find the exponential growth function that models the popula-
tion of the island. 2. Predict the population of the island in the year
2023.
Step-by-Step Solution:
Step 1: Use given data to find the growth rate k.
Given: - P(0) = 1500 (Population at the year 2010) - P(5) = 2450
(Population at the year 2015, where t= 5 since 2015 - 2010 = 5)
The equation is:
2450 = 1500 ·e5k
Firstly solve for e5k:
e5k=2450
1500 ≈1.6333
Now, solve for k:
5k= ln(1.6333)
k=ln(1.6333)
5≈0.4908
5= 0.09816
Step 2: Write the growth function.
Using the initial population and the growth rate, the function
becomes:
P(t) = 1500 ·e0.09816t
Step 3: Predict the population for the year 2023.
To find the population in 2023: - t= 2023 −2010 = 13
2
Substitute t= 13 into the population function:
P(13) = 1500 ·e0.09816×13
P(13) = 1500 ·e1.27608
P(13) = 1500 ·3.582
P(13) ≈5373
Conclusion:
The population model for the island is P(t) = 1500 ·e0.09816t, and the
predicted population in the year 2023 is approximately 5,373.
Note: This question provides a comprehensive start to under-
standing exponential functions and their applications in real-world
scenarios. It encourages students to manipulate exponential equa-
tions and use logarithms, which are part of the curriculum at Liberty
University.
Question 2
Scenario: A biologist at Liberty University is studying a popu-
lation of bacteria that doubles in size every 3 hours. Initially, the
bacteria culture has a population of 150 bacteria.
Task: Calculate the population of the bacteria after 24 hours.
Questions: 1. What is the initial population of bacteria? 2. What
is the doubling time of the bacterial population in hours? 3. Use the
formula for exponential growth to find the population of the bacteria
after 24 hours. 4. What will be the population after 9 hours?
—
Solutions:
Step 1: Understand the initial setup.
1. Question: What is the initial population of bacteria? Answer:
The initial population of bacteria is 150.
Step 2: Understand the growth pattern.
2. Question: What is the doubling time of the bacterial population
in hours? Answer: The doubling time of the bacterial population is
3 hours.
Step 3: Apply exponential growth formula.
The exponential growth formula is:
P(t) = P0×2t/T
where: - P(t)is the population at time t, - P0is the initial population,
-Tis the doubling time, - tis the elapsed time in hours.
3. Question: Use the formula for exponential growth to find the
population of the at 24 hours. Solution: Substitute the values into
3
the formula: - P0= 150 (initial population), - T= 3 (doubling time), -
t= 24 (time after which the population is needed).
Calculation:
P(24) = 150 ×224/3= 150 ×28= 150 ×256 = 38400
Answer: The population of the bacteria after 24 hours is 38,400.
Step 4: Calculate the population after another specified time.
4. Question: What will be the population after 9 hours? Solution:
Again, use the exponential growth formula:
P(9) = 150 ×29/3= 150 ×23= 150 ×8 = 1200
Answer: The population of the bacteria after 9 hours will be 1200.
These calculations provide a clear practical application of expo-
nential growth in biological studies, useful for academic purposes at
Liberty University. Question 2: Exponential Growth Model
Scenario: A biologist at Liberty University is studying a popu-
lation of bacteria that doubles in size every 3 hours. Initially, the
bacteria culture has a population of 150 bacteria.
Task: Calculate the population of the bacteria after 24 hours.
Questions: 1. What is the initial population of bacteria? 2. What
is the doubling time of the bacterial population in hours? 3. Use the
formula for exponential growth to find the population of the bacteria
after 24 hours. 4. What will be the population after 9 hours?
—
Solutions:
Step 1: Understand the initial setup.
1. Question: What is the initial population of bacteria? Answer:
The initial population of bacteria is 150.
Step 2: Understand the growth pattern.
2. Question: What is the doubling time of the bacterial population
in hours? Answer: The doubling time of the bacterial population is
3 hours.
Step 3: Apply exponential growth formula.
The exponential growth formula is:
P(t) = P0×2t/T
where: - P(t)is the population at time t, - P0is the initial population,
-Tis the doubling time, - tis the elapsed time in hours.
3. Question: Use the formula for exponential growth to find the
population of the at 24 hours. Solution: Substitute the values into
the formula: - P0= 150 (initial population), - T= 3 (doubling time), -
t= 24 (time after which the population is needed).
Calculation:
P(24) = 150 ×224/3= 150 ×28= 150 ×256 = 38400
4
Answer: The population of the bacteria after 24 hours is 38,400.
Step 4: Calculate the population after another specified time.
4. Question: What will be the population after 9 hours? Solution:
Again, use the exponential growth formula:
P(9) = 150 ×29/3= 150 ×23= 150 ×8 = 1200
Answer: The population of the bacteria after 9 hours will be 1200.
These calculations provide a clear practical application of expo-
nential growth in biological studies, useful for academic purposes at
Liberty University.
Question 3
Liberty University’s Biology Department is studying a population
of bacteria that decreases in number over time due to a toxic agent.
The initial count of bacteria is 1,000,000. The number of bacteria
decreases by 8
a) What is the formula that represents the number of bacteria,
N(t), after thours?
b) Calculate the number of bacteria remaining after 6 hours.
c) Determine how long it will take for the bacteria population to
reduce to below 100,000.
—
Solutions
Part (a): Exponential Decay Formula
Step 1: Recall the formula for exponential decay:
N(t) = N0·e−kt
where: - N0is the initial amount. - kis the decay constant. - tis the
time.
Step 2: Since the bacteria decrease by 8
Step 3: Rewrite the decay model using the decay factor:
N(t) = 1000000 ·(0.92)t
This formula will allow us to calculate the number of bacteria
remaining after any given time t.
Part (b): Bacteria Remaining after 6 Hours
Step 1: Use the formula from part (a) with t= 6.
N(6) = 1000000 ·(0.92)6
Step 2: Calculate (0.92)6using a calculator:
(0.92)6≈0.6302
5
Step 3: Now, multiply by the initial number of bacteria:
N(6) ≈1000000 ·0.6302 = 630200
So, approximately 630,200 bacteria remain after 6 hours.
Part (c): Time to Reduce to Below 100,000 Bacteria
Step 1: Set up the inequality using the decay formula:
N(t)<100000
1000000 ·(0.92)t<100000
Step 2: Divide both sides by 1,000,000:
(0.92)t<0.1
Step 3: Take the natural logarithm of both sides to solve for t:
ln((0.92)t)<ln(0.1)
t·ln(0.92) <ln(0.1)
Step 4: Calculate using a calculator:
ln(0.92) ≈ −0.0834
ln(0.1) ≈ −2.3026
t > −2.3026
−0.0834 ≈27.6
Therefore, it will take approximately 27.6 hours for the population
to reduce to below 100,000.
Conclusion
This set of calculations allows the Biology Department at Liberty
University to understand and predict the behavior of the bacteria
population under the influence of the toxic agent over time. Question
3: Exponential Decay Model
Liberty University’s Biology Department is studying a population
of bacteria that decreases in number over time due to a toxic agent.
The initial count of bacteria is 1,000,000. The number of bacteria
decreases by 8
a) What is the formula that represents the number of bacteria,
N(t), after thours?
b) Calculate the number of bacteria remaining after 6 hours.
c) Determine how long it will take for the bacteria population to
reduce to below 100,000.
—
Solutions
Part (a): Exponential Decay Formula
6
Step 1: Recall the formula for exponential decay:
N(t) = N0·e−kt
where: - N0is the initial amount. - kis the decay constant. - tis the
time.
Step 2: Since the bacteria decrease by 8
Step 3: Rewrite the decay model using the decay factor:
N(t) = 1000000 ·(0.92)t
This formula will allow us to calculate the number of bacteria
remaining after any given time t.
Part (b): Bacteria Remaining after 6 Hours
Step 1: Use the formula from part (a) with t= 6.
N(6) = 1000000 ·(0.92)6
Step 2: Calculate (0.92)6using a calculator:
(0.92)6≈0.6302
Step 3: Now, multiply by the initial number of bacteria:
N(6) ≈1000000 ·0.6302 = 630200
So, approximately 630,200 bacteria remain after 6 hours.
Part (c): Time to Reduce to Below 100,000 Bacteria
Step 1: Set up the inequality using the decay formula:
N(t)<100000
1000000 ·(0.92)t<100000
Step 2: Divide both sides by 1,000,000:
(0.92)t<0.1
Step 3: Take the natural logarithm of both sides to solve for t:
ln((0.92)t)<ln(0.1)
t·ln(0.92) <ln(0.1)
Step 4: Calculate using a calculator:
ln(0.92) ≈ −0.0834
ln(0.1) ≈ −2.3026
t > −2.3026
−0.0834 ≈27.6
Therefore, it will take approximately 27.6 hours for the population
to reduce to below 100,000.
Conclusion
This set of calculations allows the Biology Department at Liberty
University to understand and predict the behavior of the bacteria
population under the influence of the toxic agent over time.
7
Question 4
Problem: Solve the exponential equation for x:
32x+1 = 81
Step-by-step Solution:
Step 1: Recognize and rewrite the constant as a power of the base.
The number 81 can be rewritten as a power of 3, because 81 = 34.
Thus, the equation becomes:
32x+1 = 34
Step 2: Set the exponents equal to each other. Since the bases
are the same and both sides of the equation are equal, we can set the
exponents equal to each other:
2x+ 1 = 4
Step 3: Solve for x. Subtract 1 from both sides:
2x= 4 −1
2x= 3
Divide both sides by 2:
x=3
2
Step 4: Check the solution. Plug x=3
2back into the original
equation to verify:
32(3
2)+1 = 33+1 = 34= 81
Since both sides of the equation are equal, our solution x=3
2is
correct.
Conclusion: The solution to the equation 32x+1 = 81 is x=3
2.
Question 4: Solving an Exponential Equation
Problem: Solve the exponential equation for x:
32x+1 = 81
Step-by-step Solution:
Step 1: Recognize and rewrite the constant as a power of the base.
The number 81 can be rewritten as a power of 3, because 81 = 34.
Thus, the equation becomes:
32x+1 = 34
Step 2: Set the exponents equal to each other. Since the bases
are the same and both sides of the equation are equal, we can set the
exponents equal to each other:
2x+ 1 = 4
8
Step 3: Solve for x. Subtract 1 from both sides:
2x= 4 −1
2x= 3
Divide both sides by 2:
x=3
2
Step 4: Check the solution. Plug x=3
2back into the original
equation to verify:
32(3
2)+1 = 33+1 = 34= 81
Since both sides of the equation are equal, our solution x=3
2is
correct.
Conclusion: The solution to the equation 32x+1 = 81 is x=3
2.
Question 5
Liberty University’s new online MBA program initially enrolled
300 students in its first year. Due to the rising popularity of the
program, the number of students enrolls has been increasing by ap-
proximately 20
Solution:
Step 1: Define the Variables - Let Prepresent the number of
students. - Let trepresent the number of years after the first year.
Step 2: Identify the Initial Condition - Initially, when t= 0,P=
300.
Step 3: Recognize the Type of Growth - The student enrollment
increases by 20
Step 4: Write the Exponential Growth Function - Exponential
growth can be described by the formula:
P(t) = P0×(1 + r)t
where: - P0is the initial quantity (300 students), - ris the growth
rate (20- tis the time in years.
Step 5: Plug in the Values into the Formula - Since we need to
determine the number of students enrolled after 5 years, plug t= 5
into the function:
P(5) = 300 ×(1 + 0.20)5
Step 6: Calculate Using the Formula
P(5) = 300 ×1.205
P(5) = 300 ×2.48832
9
(using a calculator for 1.205)
P(5) = 746.496
Step 7: Round to the Nearest Whole Number - Since the number
of students cannot be a fraction:
P(5) ≈746
students.
Conclusion: After 5 years, the program is projected to have ap-
proximately 746 students enrolled. Question 5: Exponential Growth
Liberty University’s new online MBA program initially enrolled
300 students in its first year. Due to the rising popularity of the
program, the number of students enrolls has been increasing by ap-
proximately 20
Solution:
Step 1: Define the Variables - Let Prepresent the number of
students. - Let trepresent the number of years after the first year.
Step 2: Identify the Initial Condition - Initially, when t= 0,P=
300.
Step 3: Recognize the Type of Growth - The student enrollment
increases by 20
Step 4: Write the Exponential Growth Function - Exponential
growth can be described by the formula:
P(t) = P0×(1 + r)t
where: - P0is the initial quantity (300 students), - ris the growth
rate (20- tis the time in years.
Step 5: Plug in the Values into the Formula - Since we need to
determine the number of students enrolled after 5 years, plug t= 5
into the function:
P(5) = 300 ×(1 + 0.20)5
Step 6: Calculate Using the Formula
P(5) = 300 ×1.205
P(5) = 300 ×2.48832
(using a calculator for 1.205)
P(5) = 746.496
Step 7: Round to the Nearest Whole Number - Since the number
of students cannot be a fraction:
P(5) ≈746
students.
Conclusion: After 5 years, the program is projected to have ap-
proximately 746 students enrolled.
10
Question 6
Problem:
Evaluate the exponential function at the given x-values: f(x) =
3x+1
a) x= 0
b) x=−2
c) x= 1
d) x=1
2
Solution:
a) Evaluate f(x) = 3x+1 for x= 0
Step 1: Substitute x= 0 into the function.
f(0) = 30+1
Step 2: Simplify the expression.
f(0) = 31= 3
Final answer: f(0) = 3
b) Evaluate f(x)=3x+1 for x=−2
Step 1: Substitute x=−2into the function.
f(−2) = 3−2+1
Step 2: Simplify the expression.
f(−2) = 3−1
Step 3: Convert the negative exponent.
f(−2) = 1
31=1
3
Final answer: f(−2) = 1
3
c) Evaluate f(x)=3x+1 for x= 1
Step 1: Substitute x= 1 into the function.
f(1) = 31+1
Step 2: Simplify the expression.
f(1) = 32= 9
Final answer: f(1) = 9
d) Evaluate f(x)=3x+1 for x=1
2
Step 1: Substitute x=1
2into the function.
f1
2= 31
2+1
11
Step 2: Simplify the expression. Recall that 31
2is the square root
of 3.
f1
2= 31
2+1 = 31.5= 3 ×30.5= 3√3
Final answer: f1
2= 3√3
These evaluations provide the specific output values based on the
input values of xwhen the function f(x)=3x+1 includes both inte-
ger and fractional inputs. Question 6: Exponents and Exponential
Functions
Problem:
Evaluate the exponential function at the given x-values: f(x) =
3x+1
a) x= 0
b) x=−2
c) x= 1
d) x=1
2
Solution:
a) Evaluate f(x) = 3x+1 for x= 0
Step 1: Substitute x= 0 into the function.
f(0) = 30+1
Step 2: Simplify the expression.
f(0) = 31= 3
Final answer: f(0) = 3
b) Evaluate f(x)=3x+1 for x=−2
Step 1: Substitute x=−2into the function.
f(−2) = 3−2+1
Step 2: Simplify the expression.
f(−2) = 3−1
Step 3: Convert the negative exponent.
f(−2) = 1
31=1
3
Final answer: f(−2) = 1
3
c) Evaluate f(x)=3x+1 for x= 1
Step 1: Substitute x= 1 into the function.
f(1) = 31+1
Step 2: Simplify the expression.
12
f(1) = 32= 9
Final answer: f(1) = 9
d) Evaluate f(x)=3x+1 for x=1
2
Step 1: Substitute x=1
2into the function.
f1
2= 31
2+1
Step 2: Simplify the expression. Recall that 31
2is the square root
of 3.
f1
2= 31
2+1 = 31.5= 3 ×30.5= 3√3
Final answer: f1
2= 3√3
These evaluations provide the specific output values based on the
input values of xwhen the function f(x) = 3x+1 includes both integer
and fractional inputs.
Question 7
Problem:
The initial amount of a radioactive substance is 400 grams. After
12 hours, the substance decays to 250 grams. Assuming that the
decay can be modeled by an exponential function, determine:
a) The decay constant for this substance. b) The amount of the
substance remaining after 24 hours. c) How long it will take for the
substance to decay to 100 grams.
Solution:
Step-by-Step Solution:
Part a: Finding the Decay Constant
We start with the general formula for exponential decay, which is:
A(t) = A0e−kt
Where: - A(t)is the amount at time t, - A0is the initial amount,
-kis the decay constant, and - tis the time.
Given: - A0= 400 grams, - A(12) = 250 grams.
1. Plug in the values into the decay formula:
250 = 400e−k×12
2. Solve for e−k×12:
e−k×12 =250
400 = 0.625
13
3. Using natural logs to solve for k:
−k×12 = ln(0.625)
k=−ln(0.625)
12
k≈−(−0.4700)
12 = 0.0392
Part b: Amount Remaining After 24 Hours
1. Now, use the decay constant kfound in part (a) to find A(24):
A(24) = 400e−0.0392×24
2. Calculate:
A(24) = 400 ·e−0.9408 ≈400 ·0.3907 ≈156.28 grams
Part c: Time to Decay to 100 Grams
1. Set A(t) = 100 grams and solve for t:
100 = 400e−0.0392t
2. Solve for e−0.0392t:
e−0.0392t=100
400 = 0.25
3. Use natural logs to solve for t:
−0.0392t= ln(0.25)
t=ln(0.25)
−0.0392 ≈−1.3863
−0.0392 ≈35.42 hours
Summary: a) The decay constant kis approximately 0.0392 per
hour. b) After 24 hours, about 156.28 grams of the substance will
remain. c) It will take approximately 35.42 hours for the substance
to decay to 100 grams. Question 7: Exponential Growth and Decay
Problem:
The initial amount of a radioactive substance is 400 grams. After
12 hours, the substance decays to 250 grams. Assuming that the
decay can be modeled by an exponential function, determine:
a) The decay constant for this substance. b) The amount of the
substance remaining after 24 hours. c) How long it will take for the
substance to decay to 100 grams.
Solution:
Step-by-Step Solution:
Part a: Finding the Decay Constant
We start with the general formula for exponential decay, which is:
14
A(t) = A0e−kt
Where: - A(t)is the amount at time t, - A0is the initial amount,
-kis the decay constant, and - tis the time.
Given: - A0= 400 grams, - A(12) = 250 grams.
1. Plug in the values into the decay formula:
250 = 400e−k×12
2. Solve for e−k×12:
e−k×12 =250
400 = 0.625
3. Using natural logs to solve for k:
−k×12 = ln(0.625)
k=−ln(0.625)
12
k≈−(−0.4700)
12 = 0.0392
Part b: Amount Remaining After 24 Hours
1. Now, use the decay constant kfound in part (a) to find A(24):
A(24) = 400e−0.0392×24
2. Calculate:
A(24) = 400 ·e−0.9408 ≈400 ·0.3907 ≈156.28 grams
Part c: Time to Decay to 100 Grams
1. Set A(t) = 100 grams and solve for t:
100 = 400e−0.0392t
2. Solve for e−0.0392t:
e−0.0392t=100
400 = 0.25
3. Use natural logs to solve for t:
−0.0392t= ln(0.25)
t=ln(0.25)
−0.0392 ≈−1.3863
−0.0392 ≈35.42 hours
Summary: a) The decay constant kis approximately 0.0392 per
hour. b) After 24 hours, about 156.28 grams of the substance will
remain. c) It will take approximately 35.42 hours for the substance
to decay to 100 grams.
15
Question 8
The population of a small town is experiencing exponential growth.
Initially, the town has a population of 2,000 people, and the rate of
growth is 8
Part A: Write the formula representing the population growth of
the town.
Part B: Calculate the population of the town in 5 years.
Part C: Determine after how many years the population of the
town will double.
Solutions:
Part A: Write the formula representing the population growth of
the country.
Step 1: Identify the initial population. - Initial population, P0=
2000 people.
Step 2: Identify the growth rate. - Annual growth rate, r= 8% =
0.08.
Step 3: Write the exponential growth formula. - The general
formula for exponential growth is P(t) = P0×ert. - For this problem,
the formula becomes:
P(t) = 2000 ×e0.08t
Part B: Calculate the population of the town in 5 years.
Step 1: Plug in the values into the exponential growth formula. -
t= 5 years. - Use the growth formula: P(5) = 2000 ×e0.08×5.
Step 2: Calculate using the exponential term. - e0.08×5=e0.4≈
1.4918 (using a scientific calculator).
Step 3: Multiply to find the population. - P(5) = 2000 ×1.4918 ≈
2983.6.
Step 4: Round to the nearest whole number (if necessary). -
Population after 5 years 2984 people.
Part C: Determine after how many years the population of the
town will double.
Step 1: Set the future population to double the initial population.
- Future population P(t)=2×2000 = 4000.
Step 2: Use the growth formula with P(t) = 4000. - 4000 = 2000 ×
e0.08t.
Step 3: Simplify and solve for t. - Divide both sides by 2000:
2 = e0.08t
- Take the natural logarithm of both sides:
ln(2) = 0.08t
-t=ln(2)
0.08 ≈0.6931
0.08 ≈8.664 years.
16
Step 4: Round to the nearest whole number (if necessary). -
Population doubles in approximately 9 years.
These steps show how to model and solve problems related to ex-
ponential growth, crucial for predicting outcomes in real-world sce-
narios such as population growth. Question 8: Exponential Growth
of a Population
The population of a small town is experiencing exponential growth.
Initially, the town has a population of 2,000 people, and the rate of
growth is 8
Part A: Write the formula representing the population growth of
the town.
Part B: Calculate the population of the town in 5 years.
Part C: Determine after how many years the population of the
town will double.
Solutions:
Part A: Write the formula representing the population growth of
the country.
Step 1: Identify the initial population. - Initial population, P0=
2000 people.
Step 2: Identify the growth rate. - Annual growth rate, r= 8% =
0.08.
Step 3: Write the exponential growth formula. - The general
formula for exponential growth is P(t) = P0×ert. - For this problem,
the formula becomes:
P(t) = 2000 ×e0.08t
Part B: Calculate the population of the town in 5 years.
Step 1: Plug in the values into the exponential growth formula. -
t= 5 years. - Use the growth formula: P(5) = 2000 ×e0.08×5.
Step 2: Calculate using the exponential term. - e0.08×5=e0.4≈
1.4918 (using a scientific calculator).
Step 3: Multiply to find the population. - P(5) = 2000 ×1.4918 ≈
2983.6.
Step 4: Round to the nearest whole number (if necessary). -
Population after 5 years 2984 people.
Part C: Determine after how many years the population of the
town will double.
Step 1: Set the future population to double the initial population.
- Future population P(t)=2×2000 = 4000.
Step 2: Use the growth formula with P(t) = 4000. - 4000 = 2000 ×
e0.08t.
Step 3: Simplify and solve for t. - Divide both sides by 2000:
2 = e0.08t
17
- Take the natural logarithm of both sides:
ln(2) = 0.08t
-t=ln(2)
0.08 ≈0.6931
0.08 ≈8.664 years.
Step 4: Round to the nearest whole number (if necessary). -
Population doubles in approximately 9 years.
These steps show how to model and solve problems related to expo-
nential growth, crucial for predicting outcomes in real-world scenarios
such as population growth.
Question 9
Liberty University is testing a new biodegradable cup. The weight
of the cup decreases exponentially over time as it degrades. Initially,
the cup weighs 30 grams. After 50 days, the cup’s weight is observed
to be 22 grams.
a) Determine the exponential decay model that describes the weight
of the cup W(t)as a function of time t, in days.
b) Predict the weight of the cup after 100 days.
c) What is the half-life of the cup, based on your model?
—
Solution:
Step 1: Understanding the Exponential Model
The general formula for exponential decay can be represented as:
W(t) = W0·ekt
where: - W(t)is the weight of the cup at time t, - W0is the initial
weight of the cup, - kis the decay constant, - eis the base of the
natural logarithm, - tis time in days.
Step 2: Using the Initial Condition
Given: - W0= 30 grams (initial weight) - W(50) = 22 grams after
50 days
We can set up the equation:
22 = 30 ·e50k
Step 3: Solving for k
1. Divide both sides by 30 to isolate the exponential term:
22
30 =e50k
2. Take the natural logarithm of both sides to solve for k:
ln 22
30= 50k
18
3. Solve for k:
k=ln 22
30
50
k=ln(0.7333)
50 ≈ −0.0103
So, the decay model is:
W(t) = 30 ·e−0.0103t
Step 4: Predicting the Weight After 100 Days
Using the model:
W(100) = 30 ·e−0.0103×100
W(100) = 30 ·e−1.03
W(100) = 30 ·0.3567 ≈10.7grams
Step 5: Determining the Half-Life
The half-life is the time it takes for the weight of the cup to halve
from its original mass:
15 = 30 ·e−0.0103t1/2
1. Divide both sides by 30:
0.5 = e−0.0103t1/2
2. Take the natural logarithm of both sides:
ln(0.5) = −0.0103t1/2
3. Solve for t1/2:
t1/2=ln(0.5)
−0.0103 ≈67.3days
Answers:
a) The exponential decay model for the weight of the cup is W(t) =
30 ·e−0.0103t.
b) The predicted weight of the cup after 100 days is approximately
10.7 grams.
c) The half-life of the cup, based on the model, is approximately
67.3 days. Question 9: Exponential Decay Model
Liberty University is testing a new biodegradable cup. The weight
of the cup decreases exponentially over time as it degrades. Initially,
the cup weighs 30 grams. After 50 days, the cup’s weight is observed
to be 22 grams.
19
a) Determine the exponential decay model that describes the weight
of the cup W(t)as a function of time t, in days.
b) Predict the weight of the cup after 100 days.
c) What is the half-life of the cup, based on your model?
—
Solution:
Step 1: Understanding the Exponential Model
The general formula for exponential decay can be represented as:
W(t) = W0·ekt
where: - W(t)is the weight of the cup at time t, - W0is the initial
weight of the cup, - kis the decay constant, - eis the base of the
natural logarithm, - tis time in days.
Step 2: Using the Initial Condition
Given: - W0= 30 grams (initial weight) - W(50) = 22 grams after
50 days
We can set up the equation:
22 = 30 ·e50k
Step 3: Solving for k
1. Divide both sides by 30 to isolate the exponential term:
22
30 =e50k
2. Take the natural logarithm of both sides to solve for k:
ln 22
30= 50k
3. Solve for k:
k=ln 22
30
50
k=ln(0.7333)
50 ≈ −0.0103
So, the decay model is:
W(t) = 30 ·e−0.0103t
Step 4: Predicting the Weight After 100 Days
Using the model:
W(100) = 30 ·e−0.0103×100
W(100) = 30 ·e−1.03
W(100) = 30 ·0.3567 ≈10.7grams
20
Step 5: Determining the Half-Life
The half-life is the time it takes for the weight of the cup to halve
from its original mass:
15 = 30 ·e−0.0103t1/2
1. Divide both sides by 30:
0.5 = e−0.0103t1/2
2. Take the natural logarithm of both sides:
ln(0.5) = −0.0103t1/2
3. Solve for t1/2:
t1/2=ln(0.5)
−0.0103 ≈67.3days
Answers:
a) The exponential decay model for the weight of the cup is W(t) =
30 ·e−0.0103t.
b) The predicted weight of the cup after 100 days is approximately
10.7 grams.
c) The half-life of the cup, based on the model, is approximately
67.3 days.
Question 10
Brandon is studying the population growth of a colony of bacteria.
At 1 p.m., the colony starts with 300 bacteria. After each hour, the
population of the colony doubles.
Calculate how many bacteria will be in the colony at 5 p.m. on
the same day.
Step-by-Step Solution:
1. Understanding the problem: - Initial population, P0, is 300
bacteria. - The population doubles every hour. Hence, the growth
rate is 100- Time from 1 p.m. to 5 p.m. is 4 hours.
2. Representing the situation with an exponential function: - The
general formula for exponential growth is P(t) = P0×(1 + r)t, where:
-P(t)= population at time t, - P0= initial population, - r= growth
rate (as a decimal), - t= time in hours. - Since the population doubles
every hour, the growth rate r= 1 (or 100
3. Setting up the equation: - In this case, the equation becomes
P(t) = 300 ×2t.
21
4. Calculate the number of bacteria at 5 p.m.: - Time elapsed
from 1 p.m. to 5 p.m. is 4 hours, so t= 4. - Plug t= 4 into the
equation:
P(4) = 300 ×24
- Calculate 24:
24= 16
- Multiply by the initial number of bacteria:
P(4) = 300 ×16 = 4800
5. Conclusion: - There will be 4,800 bacteria in the colony at 5
p.m.
Thus, using the exponential growth model, we calculated that the
bacteria population will reach 4,800 by 5 p.m. on the same day.
Question 10: Exponential Growth Model
Brandon is studying the population growth of a colony of bacteria.
At 1 p.m., the colony starts with 300 bacteria. After each hour, the
population of the colony doubles.
Calculate how many bacteria will be in the colony at 5 p.m. on
the same day.
Step-by-Step Solution:
1. Understanding the problem: - Initial population, P0, is 300
bacteria. - The population doubles every hour. Hence, the growth
rate is 100- Time from 1 p.m. to 5 p.m. is 4 hours.
2. Representing the situation with an exponential function: - The
general formula for exponential growth is P(t) = P0×(1 + r)t, where:
-P(t)= population at time t, - P0= initial population, - r= growth
rate (as a decimal), - t= time in hours. - Since the population doubles
every hour, the growth rate r= 1 (or 100
3. Setting up the equation: - In this case, the equation becomes
P(t) = 300 ×2t.
4. Calculate the number of bacteria at 5 p.m.: - Time elapsed
from 1 p.m. to 5 p.m. is 4 hours, so t= 4. - Plug t= 4 into the
equation:
P(4) = 300 ×24
- Calculate 24:
24= 16
- Multiply by the initial number of bacteria:
P(4) = 300 ×16 = 4800
5. Conclusion: - There will be 4,800 bacteria in the colony at 5
p.m.
Thus, using the exponential growth model, we calculated that the
bacteria population will reach 4,800 by 5 p.m. on the same day.
22
Substitute t= 13 into the population function:
P(13) = 1500 ·e0.09816×13
P(13) = 1500 ·e1.27608
P(13) = 1500 ·3.582
P(13) ≈5373
Conclusion:
The population model for the island is P(t) = 1500 ·e0.09816t, and the
predicted population in the year 2023 is approximately 5,373.
Note: This question provides a comprehensive start to under-
standing exponential functions and their applications in real-world
scenarios. It encourages students to manipulate exponential equa-
tions and use logarithms, which are part of the curriculum at Liberty
University.
Question 2
Scenario: A biologist at Liberty University is studying a popu-
lation of bacteria that doubles in size every 3 hours. Initially, the
bacteria culture has a population of 150 bacteria.
Task: Calculate the population of the bacteria after 24 hours.
Questions: 1. What is the initial population of bacteria? 2. What
is the doubling time of the bacterial population in hours? 3. Use the
formula for exponential growth to find the population of the bacteria
after 24 hours. 4. What will be the population after 9 hours?
—
Solutions:
Step 1: Understand the initial setup.
1. Question: What is the initial population of bacteria? Answer:
The initial population of bacteria is 150.
Step 2: Understand the growth pattern.
2. Question: What is the doubling time of the bacterial population
in hours? Answer: The doubling time of the bacterial population is
3 hours.
Step 3: Apply exponential growth formula.
The exponential growth formula is:
P(t) = P0×2t/T
where: - P(t)is the population at time t, - P0is the initial population,
-Tis the doubling time, - tis the elapsed time in hours.
3. Question: Use the formula for exponential growth to find the
population of the at 24 hours. Solution: Substitute the values into
3
the formula: - P0= 150 (initial population), - T= 3 (doubling time), -
t= 24 (time after which the population is needed).
Calculation:
P(24) = 150 ×224/3= 150 ×28= 150 ×256 = 38400
Answer: The population of the bacteria after 24 hours is 38,400.
Step 4: Calculate the population after another specified time.
4. Question: What will be the population after 9 hours? Solution:
Again, use the exponential growth formula:
P(9) = 150 ×29/3= 150 ×23= 150 ×8 = 1200
Answer: The population of the bacteria after 9 hours will be 1200.
These calculations provide a clear practical application of expo-
nential growth in biological studies, useful for academic purposes at
Liberty University. Question 2: Exponential Growth Model
Scenario: A biologist at Liberty University is studying a popu-
lation of bacteria that doubles in size every 3 hours. Initially, the
bacteria culture has a population of 150 bacteria.
Task: Calculate the population of the bacteria after 24 hours.
Questions: 1. What is the initial population of bacteria? 2. What
is the doubling time of the bacterial population in hours? 3. Use the
formula for exponential growth to find the population of the bacteria
after 24 hours. 4. What will be the population after 9 hours?
—
Solutions:
Step 1: Understand the initial setup.
1. Question: What is the initial population of bacteria? Answer:
The initial population of bacteria is 150.
Step 2: Understand the growth pattern.
2. Question: What is the doubling time of the bacterial population
in hours? Answer: The doubling time of the bacterial population is
3 hours.
Step 3: Apply exponential growth formula.
The exponential growth formula is:
P(t) = P0×2t/T
where: - P(t)is the population at time t, - P0is the initial population,
-Tis the doubling time, - tis the elapsed time in hours.
3. Question: Use the formula for exponential growth to find the
population of the at 24 hours. Solution: Substitute the values into
the formula: - P0= 150 (initial population), - T= 3 (doubling time), -
t= 24 (time after which the population is needed).
Calculation:
P(24) = 150 ×224/3= 150 ×28= 150 ×256 = 38400
4
Answer: The population of the bacteria after 24 hours is 38,400.
Step 4: Calculate the population after another specified time.
4. Question: What will be the population after 9 hours? Solution:
Again, use the exponential growth formula:
P(9) = 150 ×29/3= 150 ×23= 150 ×8 = 1200
Answer: The population of the bacteria after 9 hours will be 1200.
These calculations provide a clear practical application of expo-
nential growth in biological studies, useful for academic purposes at
Liberty University.
Question 3
Liberty University’s Biology Department is studying a population
of bacteria that decreases in number over time due to a toxic agent.
The initial count of bacteria is 1,000,000. The number of bacteria
decreases by 8
a) What is the formula that represents the number of bacteria,
N(t), after thours?
b) Calculate the number of bacteria remaining after 6 hours.
c) Determine how long it will take for the bacteria population to
reduce to below 100,000.
—
Solutions
Part (a): Exponential Decay Formula
Step 1: Recall the formula for exponential decay:
N(t) = N0·e−kt
where: - N0is the initial amount. - kis the decay constant. - tis the
time.
Step 2: Since the bacteria decrease by 8
Step 3: Rewrite the decay model using the decay factor:
N(t) = 1000000 ·(0.92)t
This formula will allow us to calculate the number of bacteria
remaining after any given time t.
Part (b): Bacteria Remaining after 6 Hours
Step 1: Use the formula from part (a) with t= 6.
N(6) = 1000000 ·(0.92)6
Step 2: Calculate (0.92)6using a calculator:
(0.92)6≈0.6302
5
Step 3: Now, multiply by the initial number of bacteria:
N(6) ≈1000000 ·0.6302 = 630200
So, approximately 630,200 bacteria remain after 6 hours.
Part (c): Time to Reduce to Below 100,000 Bacteria
Step 1: Set up the inequality using the decay formula:
N(t)<100000
1000000 ·(0.92)t<100000
Step 2: Divide both sides by 1,000,000:
(0.92)t<0.1
Step 3: Take the natural logarithm of both sides to solve for t:
ln((0.92)t)<ln(0.1)
t·ln(0.92) <ln(0.1)
Step 4: Calculate using a calculator:
ln(0.92) ≈ −0.0834
ln(0.1) ≈ −2.3026
t > −2.3026
−0.0834 ≈27.6
Therefore, it will take approximately 27.6 hours for the population
to reduce to below 100,000.
Conclusion
This set of calculations allows the Biology Department at Liberty
University to understand and predict the behavior of the bacteria
population under the influence of the toxic agent over time. Question
3: Exponential Decay Model
Liberty University’s Biology Department is studying a population
of bacteria that decreases in number over time due to a toxic agent.
The initial count of bacteria is 1,000,000. The number of bacteria
decreases by 8
a) What is the formula that represents the number of bacteria,
N(t), after thours?
b) Calculate the number of bacteria remaining after 6 hours.
c) Determine how long it will take for the bacteria population to
reduce to below 100,000.
—
Solutions
Part (a): Exponential Decay Formula
6
Step 1: Recall the formula for exponential decay:
N(t) = N0·e−kt
where: - N0is the initial amount. - kis the decay constant. - tis the
time.
Step 2: Since the bacteria decrease by 8
Step 3: Rewrite the decay model using the decay factor:
N(t) = 1000000 ·(0.92)t
This formula will allow us to calculate the number of bacteria
remaining after any given time t.
Part (b): Bacteria Remaining after 6 Hours
Step 1: Use the formula from part (a) with t= 6.
N(6) = 1000000 ·(0.92)6
Step 2: Calculate (0.92)6using a calculator:
(0.92)6≈0.6302
Step 3: Now, multiply by the initial number of bacteria:
N(6) ≈1000000 ·0.6302 = 630200
So, approximately 630,200 bacteria remain after 6 hours.
Part (c): Time to Reduce to Below 100,000 Bacteria
Step 1: Set up the inequality using the decay formula:
N(t)<100000
1000000 ·(0.92)t<100000
Step 2: Divide both sides by 1,000,000:
(0.92)t<0.1
Step 3: Take the natural logarithm of both sides to solve for t:
ln((0.92)t)<ln(0.1)
t·ln(0.92) <ln(0.1)
Step 4: Calculate using a calculator:
ln(0.92) ≈ −0.0834
ln(0.1) ≈ −2.3026
t > −2.3026
−0.0834 ≈27.6
Therefore, it will take approximately 27.6 hours for the population
to reduce to below 100,000.
Conclusion
This set of calculations allows the Biology Department at Liberty
University to understand and predict the behavior of the bacteria
population under the influence of the toxic agent over time.
7
Question 4
Problem: Solve the exponential equation for x:
32x+1 = 81
Step-by-step Solution:
Step 1: Recognize and rewrite the constant as a power of the base.
The number 81 can be rewritten as a power of 3, because 81 = 34.
Thus, the equation becomes:
32x+1 = 34
Step 2: Set the exponents equal to each other. Since the bases
are the same and both sides of the equation are equal, we can set the
exponents equal to each other:
2x+ 1 = 4
Step 3: Solve for x. Subtract 1 from both sides:
2x= 4 −1
2x= 3
Divide both sides by 2:
x=3
2
Step 4: Check the solution. Plug x=3
2back into the original
equation to verify:
32(3
2)+1 = 33+1 = 34= 81
Since both sides of the equation are equal, our solution x=3
2is
correct.
Conclusion: The solution to the equation 32x+1 = 81 is x=3
2.
Question 4: Solving an Exponential Equation
Problem: Solve the exponential equation for x:
32x+1 = 81
Step-by-step Solution:
Step 1: Recognize and rewrite the constant as a power of the base.
The number 81 can be rewritten as a power of 3, because 81 = 34.
Thus, the equation becomes:
32x+1 = 34
Step 2: Set the exponents equal to each other. Since the bases
are the same and both sides of the equation are equal, we can set the
exponents equal to each other:
2x+ 1 = 4
8
Step 3: Solve for x. Subtract 1 from both sides:
2x= 4 −1
2x= 3
Divide both sides by 2:
x=3
2
Step 4: Check the solution. Plug x=3
2back into the original
equation to verify:
32(3
2)+1 = 33+1 = 34= 81
Since both sides of the equation are equal, our solution x=3
2is
correct.
Conclusion: The solution to the equation 32x+1 = 81 is x=3
2.
Question 5
Liberty University’s new online MBA program initially enrolled
300 students in its first year. Due to the rising popularity of the
program, the number of students enrolls has been increasing by ap-
proximately 20
Solution:
Step 1: Define the Variables - Let Prepresent the number of
students. - Let trepresent the number of years after the first year.
Step 2: Identify the Initial Condition - Initially, when t= 0,P=
300.
Step 3: Recognize the Type of Growth - The student enrollment
increases by 20
Step 4: Write the Exponential Growth Function - Exponential
growth can be described by the formula:
P(t) = P0×(1 + r)t
where: - P0is the initial quantity (300 students), - ris the growth
rate (20- tis the time in years.
Step 5: Plug in the Values into the Formula - Since we need to
determine the number of students enrolled after 5 years, plug t= 5
into the function:
P(5) = 300 ×(1 + 0.20)5
Step 6: Calculate Using the Formula
P(5) = 300 ×1.205
P(5) = 300 ×2.48832
9
(using a calculator for 1.205)
P(5) = 746.496
Step 7: Round to the Nearest Whole Number - Since the number
of students cannot be a fraction:
P(5) ≈746
students.
Conclusion: After 5 years, the program is projected to have ap-
proximately 746 students enrolled. Question 5: Exponential Growth
Liberty University’s new online MBA program initially enrolled
300 students in its first year. Due to the rising popularity of the
program, the number of students enrolls has been increasing by ap-
proximately 20
Solution:
Step 1: Define the Variables - Let Prepresent the number of
students. - Let trepresent the number of years after the first year.
Step 2: Identify the Initial Condition - Initially, when t= 0,P=
300.
Step 3: Recognize the Type of Growth - The student enrollment
increases by 20
Step 4: Write the Exponential Growth Function - Exponential
growth can be described by the formula:
P(t) = P0×(1 + r)t
where: - P0is the initial quantity (300 students), - ris the growth
rate (20- tis the time in years.
Step 5: Plug in the Values into the Formula - Since we need to
determine the number of students enrolled after 5 years, plug t= 5
into the function:
P(5) = 300 ×(1 + 0.20)5
Step 6: Calculate Using the Formula
P(5) = 300 ×1.205
P(5) = 300 ×2.48832
(using a calculator for 1.205)
P(5) = 746.496
Step 7: Round to the Nearest Whole Number - Since the number
of students cannot be a fraction:
P(5) ≈746
students.
Conclusion: After 5 years, the program is projected to have ap-
proximately 746 students enrolled.
10
Question 6
Problem:
Evaluate the exponential function at the given x-values: f(x) =
3x+1
a) x= 0
b) x=−2
c) x= 1
d) x=1
2
Solution:
a) Evaluate f(x) = 3x+1 for x= 0
Step 1: Substitute x= 0 into the function.
f(0) = 30+1
Step 2: Simplify the expression.
f(0) = 31= 3
Final answer: f(0) = 3
b) Evaluate f(x)=3x+1 for x=−2
Step 1: Substitute x=−2into the function.
f(−2) = 3−2+1
Step 2: Simplify the expression.
f(−2) = 3−1
Step 3: Convert the negative exponent.
f(−2) = 1
31=1
3
Final answer: f(−2) = 1
3
c) Evaluate f(x)=3x+1 for x= 1
Step 1: Substitute x= 1 into the function.
f(1) = 31+1
Step 2: Simplify the expression.
f(1) = 32= 9
Final answer: f(1) = 9
d) Evaluate f(x)=3x+1 for x=1
2
Step 1: Substitute x=1
2into the function.
f1
2= 31
2+1
11
Step 2: Simplify the expression. Recall that 31
2is the square root
of 3.
f1
2= 31
2+1 = 31.5= 3 ×30.5= 3√3
Final answer: f1
2= 3√3
These evaluations provide the specific output values based on the
input values of xwhen the function f(x)=3x+1 includes both inte-
ger and fractional inputs. Question 6: Exponents and Exponential
Functions
Problem:
Evaluate the exponential function at the given x-values: f(x) =
3x+1
a) x= 0
b) x=−2
c) x= 1
d) x=1
2
Solution:
a) Evaluate f(x) = 3x+1 for x= 0
Step 1: Substitute x= 0 into the function.
f(0) = 30+1
Step 2: Simplify the expression.
f(0) = 31= 3
Final answer: f(0) = 3
b) Evaluate f(x)=3x+1 for x=−2
Step 1: Substitute x=−2into the function.
f(−2) = 3−2+1
Step 2: Simplify the expression.
f(−2) = 3−1
Step 3: Convert the negative exponent.
f(−2) = 1
31=1
3
Final answer: f(−2) = 1
3
c) Evaluate f(x)=3x+1 for x= 1
Step 1: Substitute x= 1 into the function.
f(1) = 31+1
Step 2: Simplify the expression.
12
f(1) = 32= 9
Final answer: f(1) = 9
d) Evaluate f(x)=3x+1 for x=1
2
Step 1: Substitute x=1
2into the function.
f1
2= 31
2+1
Step 2: Simplify the expression. Recall that 31
2is the square root
of 3.
f1
2= 31
2+1 = 31.5= 3 ×30.5= 3√3
Final answer: f1
2= 3√3
These evaluations provide the specific output values based on the
input values of xwhen the function f(x) = 3x+1 includes both integer
and fractional inputs.
Question 7
Problem:
The initial amount of a radioactive substance is 400 grams. After
12 hours, the substance decays to 250 grams. Assuming that the
decay can be modeled by an exponential function, determine:
a) The decay constant for this substance. b) The amount of the
substance remaining after 24 hours. c) How long it will take for the
substance to decay to 100 grams.
Solution:
Step-by-Step Solution:
Part a: Finding the Decay Constant
We start with the general formula for exponential decay, which is:
A(t) = A0e−kt
Where: - A(t)is the amount at time t, - A0is the initial amount,
-kis the decay constant, and - tis the time.
Given: - A0= 400 grams, - A(12) = 250 grams.
1. Plug in the values into the decay formula:
250 = 400e−k×12
2. Solve for e−k×12:
e−k×12 =250
400 = 0.625
13
3. Using natural logs to solve for k:
−k×12 = ln(0.625)
k=−ln(0.625)
12
k≈−(−0.4700)
12 = 0.0392
Part b: Amount Remaining After 24 Hours
1. Now, use the decay constant kfound in part (a) to find A(24):
A(24) = 400e−0.0392×24
2. Calculate:
A(24) = 400 ·e−0.9408 ≈400 ·0.3907 ≈156.28 grams
Part c: Time to Decay to 100 Grams
1. Set A(t) = 100 grams and solve for t:
100 = 400e−0.0392t
2. Solve for e−0.0392t:
e−0.0392t=100
400 = 0.25
3. Use natural logs to solve for t:
−0.0392t= ln(0.25)
t=ln(0.25)
−0.0392 ≈−1.3863
−0.0392 ≈35.42 hours
Summary: a) The decay constant kis approximately 0.0392 per
hour. b) After 24 hours, about 156.28 grams of the substance will
remain. c) It will take approximately 35.42 hours for the substance
to decay to 100 grams. Question 7: Exponential Growth and Decay
Problem:
The initial amount of a radioactive substance is 400 grams. After
12 hours, the substance decays to 250 grams. Assuming that the
decay can be modeled by an exponential function, determine:
a) The decay constant for this substance. b) The amount of the
substance remaining after 24 hours. c) How long it will take for the
substance to decay to 100 grams.
Solution:
Step-by-Step Solution:
Part a: Finding the Decay Constant
We start with the general formula for exponential decay, which is:
14
A(t) = A0e−kt
Where: - A(t)is the amount at time t, - A0is the initial amount,
-kis the decay constant, and - tis the time.
Given: - A0= 400 grams, - A(12) = 250 grams.
1. Plug in the values into the decay formula:
250 = 400e−k×12
2. Solve for e−k×12:
e−k×12 =250
400 = 0.625
3. Using natural logs to solve for k:
−k×12 = ln(0.625)
k=−ln(0.625)
12
k≈−(−0.4700)
12 = 0.0392
Part b: Amount Remaining After 24 Hours
1. Now, use the decay constant kfound in part (a) to find A(24):
A(24) = 400e−0.0392×24
2. Calculate:
A(24) = 400 ·e−0.9408 ≈400 ·0.3907 ≈156.28 grams
Part c: Time to Decay to 100 Grams
1. Set A(t) = 100 grams and solve for t:
100 = 400e−0.0392t
2. Solve for e−0.0392t:
e−0.0392t=100
400 = 0.25
3. Use natural logs to solve for t:
−0.0392t= ln(0.25)
t=ln(0.25)
−0.0392 ≈−1.3863
−0.0392 ≈35.42 hours
Summary: a) The decay constant kis approximately 0.0392 per
hour. b) After 24 hours, about 156.28 grams of the substance will
remain. c) It will take approximately 35.42 hours for the substance
to decay to 100 grams.
15
Question 8
The population of a small town is experiencing exponential growth.
Initially, the town has a population of 2,000 people, and the rate of
growth is 8
Part A: Write the formula representing the population growth of
the town.
Part B: Calculate the population of the town in 5 years.
Part C: Determine after how many years the population of the
town will double.
Solutions:
Part A: Write the formula representing the population growth of
the country.
Step 1: Identify the initial population. - Initial population, P0=
2000 people.
Step 2: Identify the growth rate. - Annual growth rate, r= 8% =
0.08.
Step 3: Write the exponential growth formula. - The general
formula for exponential growth is P(t) = P0×ert. - For this problem,
the formula becomes:
P(t) = 2000 ×e0.08t
Part B: Calculate the population of the town in 5 years.
Step 1: Plug in the values into the exponential growth formula. -
t= 5 years. - Use the growth formula: P(5) = 2000 ×e0.08×5.
Step 2: Calculate using the exponential term. - e0.08×5=e0.4≈
1.4918 (using a scientific calculator).
Step 3: Multiply to find the population. - P(5) = 2000 ×1.4918 ≈
2983.6.
Step 4: Round to the nearest whole number (if necessary). -
Population after 5 years 2984 people.
Part C: Determine after how many years the population of the
town will double.
Step 1: Set the future population to double the initial population.
- Future population P(t)=2×2000 = 4000.
Step 2: Use the growth formula with P(t) = 4000. - 4000 = 2000 ×
e0.08t.
Step 3: Simplify and solve for t. - Divide both sides by 2000:
2 = e0.08t
- Take the natural logarithm of both sides:
ln(2) = 0.08t
-t=ln(2)
0.08 ≈0.6931
0.08 ≈8.664 years.
16
Step 4: Round to the nearest whole number (if necessary). -
Population doubles in approximately 9 years.
These steps show how to model and solve problems related to ex-
ponential growth, crucial for predicting outcomes in real-world sce-
narios such as population growth. Question 8: Exponential Growth
of a Population
The population of a small town is experiencing exponential growth.
Initially, the town has a population of 2,000 people, and the rate of
growth is 8
Part A: Write the formula representing the population growth of
the town.
Part B: Calculate the population of the town in 5 years.
Part C: Determine after how many years the population of the
town will double.
Solutions:
Part A: Write the formula representing the population growth of
the country.
Step 1: Identify the initial population. - Initial population, P0=
2000 people.
Step 2: Identify the growth rate. - Annual growth rate, r= 8% =
0.08.
Step 3: Write the exponential growth formula. - The general
formula for exponential growth is P(t) = P0×ert. - For this problem,
the formula becomes:
P(t) = 2000 ×e0.08t
Part B: Calculate the population of the town in 5 years.
Step 1: Plug in the values into the exponential growth formula. -
t= 5 years. - Use the growth formula: P(5) = 2000 ×e0.08×5.
Step 2: Calculate using the exponential term. - e0.08×5=e0.4≈
1.4918 (using a scientific calculator).
Step 3: Multiply to find the population. - P(5) = 2000 ×1.4918 ≈
2983.6.
Step 4: Round to the nearest whole number (if necessary). -
Population after 5 years 2984 people.
Part C: Determine after how many years the population of the
town will double.
Step 1: Set the future population to double the initial population.
- Future population P(t)=2×2000 = 4000.
Step 2: Use the growth formula with P(t) = 4000. - 4000 = 2000 ×
e0.08t.
Step 3: Simplify and solve for t. - Divide both sides by 2000:
2 = e0.08t
17
- Take the natural logarithm of both sides:
ln(2) = 0.08t
-t=ln(2)
0.08 ≈0.6931
0.08 ≈8.664 years.
Step 4: Round to the nearest whole number (if necessary). -
Population doubles in approximately 9 years.
These steps show how to model and solve problems related to expo-
nential growth, crucial for predicting outcomes in real-world scenarios
such as population growth.
Question 9
Liberty University is testing a new biodegradable cup. The weight
of the cup decreases exponentially over time as it degrades. Initially,
the cup weighs 30 grams. After 50 days, the cup’s weight is observed
to be 22 grams.
a) Determine the exponential decay model that describes the weight
of the cup W(t)as a function of time t, in days.
b) Predict the weight of the cup after 100 days.
c) What is the half-life of the cup, based on your model?
—
Solution:
Step 1: Understanding the Exponential Model
The general formula for exponential decay can be represented as:
W(t) = W0·ekt
where: - W(t)is the weight of the cup at time t, - W0is the initial
weight of the cup, - kis the decay constant, - eis the base of the
natural logarithm, - tis time in days.
Step 2: Using the Initial Condition
Given: - W0= 30 grams (initial weight) - W(50) = 22 grams after
50 days
We can set up the equation:
22 = 30 ·e50k
Step 3: Solving for k
1. Divide both sides by 30 to isolate the exponential term:
22
30 =e50k
2. Take the natural logarithm of both sides to solve for k:
ln 22
30= 50k
18
3. Solve for k:
k=ln 22
30
50
k=ln(0.7333)
50 ≈ −0.0103
So, the decay model is:
W(t) = 30 ·e−0.0103t
Step 4: Predicting the Weight After 100 Days
Using the model:
W(100) = 30 ·e−0.0103×100
W(100) = 30 ·e−1.03
W(100) = 30 ·0.3567 ≈10.7grams
Step 5: Determining the Half-Life
The half-life is the time it takes for the weight of the cup to halve
from its original mass:
15 = 30 ·e−0.0103t1/2
1. Divide both sides by 30:
0.5 = e−0.0103t1/2
2. Take the natural logarithm of both sides:
ln(0.5) = −0.0103t1/2
3. Solve for t1/2:
t1/2=ln(0.5)
−0.0103 ≈67.3days
Answers:
a) The exponential decay model for the weight of the cup is W(t) =
30 ·e−0.0103t.
b) The predicted weight of the cup after 100 days is approximately
10.7 grams.
c) The half-life of the cup, based on the model, is approximately
67.3 days. Question 9: Exponential Decay Model
Liberty University is testing a new biodegradable cup. The weight
of the cup decreases exponentially over time as it degrades. Initially,
the cup weighs 30 grams. After 50 days, the cup’s weight is observed
to be 22 grams.
19
a) Determine the exponential decay model that describes the weight
of the cup W(t)as a function of time t, in days.
b) Predict the weight of the cup after 100 days.
c) What is the half-life of the cup, based on your model?
—
Solution:
Step 1: Understanding the Exponential Model
The general formula for exponential decay can be represented as:
W(t) = W0·ekt
where: - W(t)is the weight of the cup at time t, - W0is the initial
weight of the cup, - kis the decay constant, - eis the base of the
natural logarithm, - tis time in days.
Step 2: Using the Initial Condition
Given: - W0= 30 grams (initial weight) - W(50) = 22 grams after
50 days
We can set up the equation:
22 = 30 ·e50k
Step 3: Solving for k
1. Divide both sides by 30 to isolate the exponential term:
22
30 =e50k
2. Take the natural logarithm of both sides to solve for k:
ln 22
30= 50k
3. Solve for k:
k=ln 22
30
50
k=ln(0.7333)
50 ≈ −0.0103
So, the decay model is:
W(t) = 30 ·e−0.0103t
Step 4: Predicting the Weight After 100 Days
Using the model:
W(100) = 30 ·e−0.0103×100
W(100) = 30 ·e−1.03
W(100) = 30 ·0.3567 ≈10.7grams
20
Step 5: Determining the Half-Life
The half-life is the time it takes for the weight of the cup to halve
from its original mass:
15 = 30 ·e−0.0103t1/2
1. Divide both sides by 30:
0.5 = e−0.0103t1/2
2. Take the natural logarithm of both sides:
ln(0.5) = −0.0103t1/2
3. Solve for t1/2:
t1/2=ln(0.5)
−0.0103 ≈67.3days
Answers:
a) The exponential decay model for the weight of the cup is W(t) =
30 ·e−0.0103t.
b) The predicted weight of the cup after 100 days is approximately
10.7 grams.
c) The half-life of the cup, based on the model, is approximately
67.3 days.
Question 10
Brandon is studying the population growth of a colony of bacteria.
At 1 p.m., the colony starts with 300 bacteria. After each hour, the
population of the colony doubles.
Calculate how many bacteria will be in the colony at 5 p.m. on
the same day.
Step-by-Step Solution:
1. Understanding the problem: - Initial population, P0, is 300
bacteria. - The population doubles every hour. Hence, the growth
rate is 100- Time from 1 p.m. to 5 p.m. is 4 hours.
2. Representing the situation with an exponential function: - The
general formula for exponential growth is P(t) = P0×(1 + r)t, where:
-P(t)= population at time t, - P0= initial population, - r= growth
rate (as a decimal), - t= time in hours. - Since the population doubles
every hour, the growth rate r= 1 (or 100
3. Setting up the equation: - In this case, the equation becomes
P(t) = 300 ×2t.
21
4. Calculate the number of bacteria at 5 p.m.: - Time elapsed
from 1 p.m. to 5 p.m. is 4 hours, so t= 4. - Plug t= 4 into the
equation:
P(4) = 300 ×24
- Calculate 24:
24= 16
- Multiply by the initial number of bacteria:
P(4) = 300 ×16 = 4800
5. Conclusion: - There will be 4,800 bacteria in the colony at 5
p.m.
Thus, using the exponential growth model, we calculated that the
bacteria population will reach 4,800 by 5 p.m. on the same day.
Question 10: Exponential Growth Model
Brandon is studying the population growth of a colony of bacteria.
At 1 p.m., the colony starts with 300 bacteria. After each hour, the
population of the colony doubles.
Calculate how many bacteria will be in the colony at 5 p.m. on
the same day.
Step-by-Step Solution:
1. Understanding the problem: - Initial population, P0, is 300
bacteria. - The population doubles every hour. Hence, the growth
rate is 100- Time from 1 p.m. to 5 p.m. is 4 hours.
2. Representing the situation with an exponential function: - The
general formula for exponential growth is P(t) = P0×(1 + r)t, where:
-P(t)= population at time t, - P0= initial population, - r= growth
rate (as a decimal), - t= time in hours. - Since the population doubles
every hour, the growth rate r= 1 (or 100
3. Setting up the equation: - In this case, the equation becomes
P(t) = 300 ×2t.
4. Calculate the number of bacteria at 5 p.m.: - Time elapsed
from 1 p.m. to 5 p.m. is 4 hours, so t= 4. - Plug t= 4 into the
equation:
P(4) = 300 ×24
- Calculate 24:
24= 16
- Multiply by the initial number of bacteria:
P(4) = 300 ×16 = 4800
5. Conclusion: - There will be 4,800 bacteria in the colony at 5
p.m.
Thus, using the exponential growth model, we calculated that the
bacteria population will reach 4,800 by 5 p.m. on the same day.
22
